Finding a particular solution of $y''-y'-2y = 4e^{-t}$ I am to find a particular solution of $$y'' - y' - 2y = 4e^{-t}$$
Guessing a solution of the form $y(t) = Ae^{-t}$, we obtain the contradiction $0 = 4$. Thus, its safe to assume that the solution is not on the form $Ae^{-t}$. I don't see the idea here. Am I supposed to construct some $Ag(t)$ that forces output of the form $Ae^{-t}$?
 A: If $D$ is the differentiation operator, your equation is
$$
                (D+1)(D-2)y=4e^{-t}
$$
And $(D+1)e^{-t}=0$. So, any particular solution will also be a solution of
$$
                    (D+1)^{2}(D-2)y=0.
$$
Any particular solution must have the form
$$
                    y = Ae^{2t}+Be^{-t}+Cte^{-t}
$$
for some constants $A$, $B$, $C$ yet to be determined. The first two terms are annihilated by $(D+1)(D-2)$. So you only have to try $Cte^{-t}$ for a particular solution. Let $w=Cte^{-t}$. Then
$$
\begin{align}
       (D+1)w & = (Ce^{-t}-Cte^{-t})+Cte^{-t}= Ce^{-t} \\
       (D-2)(D+1)w & = (D-2)(Ce^{-t})=(-C-2C)e^{-t}=-3Ce^{-t}
\end{align}
$$
So you want $C=-4/3$. That is, $-\frac{4}{3}te^{-t}$ is a particular solution, and the general solution is $Ae^{2t}+Be^{-t}-\frac{4}{3}te^{-t}$.
A: Much less clever than T.A.E., using the standard procedure of variation of parameters, the general solution of the ODE is $$y(t)=c_1 e^{-t}+c_2 e^{2 t}-\frac{4}{9} e^{-t} (3 t+1)  $$ and, in my opinion, this is the first step. Now, you can constraint the solution playing with coefficients $c_1$ and $c_2$. 
A: This is a relatively easy problem to use integrating factors.  Since the roots of the characteristic polynomial are $-1$ or $2$, integrating factors of $e^t$ or $e^{-2t}$ will work.  $e^{-2t}$ will make the second integration easier.  So, we have
$$(y'+y)'-2(y'+y)=4e^{-t}$$
$$e^{-2t}(y'+y)'-2e^{-2t}(y'+y)=[e^{-2t}(y'+y)]'=4e^{-3t}$$
$$e^{-2t}(y'+y)=-\frac43e^{-3t}+k_1$$
$$e^ty'+e^ty=(e^ty)'=-\frac43+k_1e^{3t}$$
$$e^ty=-\frac43t+k_2e^{3t}+k_3$$
$$y=-\frac43te^{-t}+k_2e^{2t}+k_3e^{-t}$$
